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a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
a)\(A=x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\). Tại \(\left|2x+1\right|=2\) thì:
\(\Rightarrow2x+1=\pm2\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
*)Xét \(x=\dfrac{1}{2}\Rightarrow A=\left(x-1\right)^3=\left(\dfrac{1}{2}-1\right)^3=-\dfrac{1}{8}\)
*)Xét \(x=-\dfrac{3}{2}\Rightarrow A=\left(x-1\right)^3=\left(-\dfrac{3}{2}-1\right)^3=-\dfrac{125}{8}\)
b)Tại \(x^2+y^2=1\) thì:
\(B=2x^4+3x^2y^2+y^4+y^2\)
\(=2x^4+2x^2y^2+x^2y^2+y^4+y^2\)
\(=2x^2\left(x^2+y^2\right)+y^2\left(x^2+y^2\right)+y^2\)
\(=2x^2+y^2+y^2=2x^2+2y^2\)
\(=2\left(x^2+y^2\right)=2\cdot1=2\)
Bài 1:
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2=x+y+z\)
+) \(\dfrac{y+z+1}{x}=2\Rightarrow y+z+1=2x\)
\(\Rightarrow x+y+z+1=3x\)
\(\Rightarrow3=3x\Rightarrow x=1\)
+) \(\dfrac{x+z+2}{y}=2\Rightarrow x+z+2=2y\)
\(\Rightarrow x+y+z+2=3y\Rightarrow y=\dfrac{4}{3}\)
+) \(\dfrac{x+y-3}{z}=2\Rightarrow x+y-3=2z\)
\(\Rightarrow x+y+z-3=3z\)
\(\Rightarrow z=\dfrac{-1}{3}\)
Vậy...
Bài 2:
Giải:
Ta có: \(\dfrac{2+3x}{4}=\dfrac{1-5x}{2}\)
\(\Rightarrow4+6x=4-20x\)
\(\Rightarrow26x=0\Rightarrow x=0\)
\(\dfrac{1-5x}{2}=\dfrac{y+2x}{2y+3x}\)
\(\Rightarrow\dfrac{1}{2}=\dfrac{y}{2y}\)
\(\Rightarrow2y=2y\)
\(\Rightarrow y\in R\left(y\ne0\right)\)
Vậy....
\(G=3x^4+5x^2y^2+2y^4+2x^2\)
\(G=3x^4+3x^2y^2+2x^2y^2+2y^4++2x^2\)
\(G=3x^2.\left(x^2+y^2\right)+2y^2.\left(x^2+y^2\right)+2x^2\)
\(G=3x^2.0+2y^2.0+2x^2\)
\(G=2x^2\)
\(M=\left(x^2+y^2\right)^2+x^4+x^2y^2+y^2\)
\(M=1+x^2\left(x^2+y^2\right)+y^2\)
\(M=1+x^2+y^2\)
\(M=1+1=2\)
câu b bạn xem lại đề ạ chắc thiếu mất dấu cộng
\(4\left(3x^2+5x+2\right)=0\Leftrightarrow12x^2+20x+8=0\)
\(\Leftrightarrow12x^2+20x+1=-7\)
thanks bn nhaa, mk k cho bn rr